36 Higher Engineering Mathematics
0
2
4
6
5.71
8
0.5
0.555
i (A)
t(s)
1.0
1.5
i 5 8.0 (12e 2t/CR )
Figure 4.6
Problem 20. The temperature θ 2 of a winding
which is being heated electrically at time t is given
by: θ 2 = θ 1 (1 − e
−t
τ ) where θ 1 is the temperature (in
degrees Celsius) at time t = 0 and τ is a constant.
Calculate,
(a) θ 1 , correct to the nearest degree, when θ 2 is
50 ◦ C, t is 30 s and τ is 60 s
(b) the time t , correct to 1 decimal place, for θ 2 to
be half the value of θ 1 .
(a) Transposing the formula to make θ 1 the subject
gives:
θ 1 =
θ 2
(1 − e
−t
T )
=
50
1 − e
−30
60
=
50
1 − e −0.5 =
50
0.393469 ...
i.e. θ 1 = 127
◦ C, correct to the nearest degree.
(b) Transposing to make t the subject of the formula
gives:
θ 2
θ 1
= 1 − e
−t
τ
from which, e
−t
τ = 1 −
θ 2
θ 1
Hence
−
t
τ
= ln
1 −
θ 2
θ 1
i.e.
t = −τ ln
1 −
θ 2
θ 1
Since
θ 2 =
1
2
θ 1
t = −60 ln
1 −
1
2
= −60 ln 0.5 = 41.59 s
Hence the time for the temperature θ 2 to be
one half of the value of θ 1 is 41.6 s, correct to 1
decimal place.
Now try the following exercise
Exercise 18 Further problems on the laws
of growth and decay
1. The temperature, T ◦ C, of a cooling object
varies with time, t minutes, according to the
equation: T = 150e −0.04t . Determine the temperature when (a) t = 0, (b) t = 10 minutes.
[(a) 150 ◦ C (b) 100.5 ◦ C ]
2. The pressure p pascals at height h metres
above ground level is given by p = p 0 e
−h
C ,
where p 0 is the pressure at ground level
and C is a constant. Find pressure p when
p 0 = 1.012 × 10 5 Pa, height h = 1420 m, and
C = 71500.
[99210]
3. The voltage drop, v volts, across an inductor L henrys at time t seconds is given
by v = 200 e
− Rt
L , where R = 150 and
L =12.5 × 10 −3 H. Determine (a) the voltage
when t = 160 ×10 −6 s, and (b) the time for the
voltage to reach 85 V.
[(a) 29.32 volts (b) 71.31 × 10 −6 s]
4. The length l metres of a metal bar at temperature t ◦ C is given by l = l 0 e αt , where
l 0 and α are constants. Determine (a) the
value of α when l = 1.993 m, l 0 = 1.894 m
and t = 250 ◦ C, and (b) the value of l 0 when
l = 2.416, t = 310 ◦ C and α = 1.682 ×10 −4 .
[(a) 2.038 × 10 −4 (b) 2.293 m]
5. The temperature θ
◦
2 C of an electrical conductor at time t seconds is given by:
θ 2 = θ 1 (1 − e −t / T ), where θ 1 is the initial
temperature and T seconds is a constant.
Determine:
(a) θ 2 when θ 1 = 159.9 ◦ C, t = 30 s and
T = 80 s, and
(b) the time t for θ 2 to fall to half the value
of θ 1 if T remains at 80 s.
[(a) 50 ◦ C (b) 55.45 s ]
6. A belt is in contact with a pulley for a
sector of θ = 1.12 radians and the coefficient
0
2
4
6
5.71
8
0.5
0.555
i (A)
t(s)
1.0
1.5
i 5 8.0 (12e 2t/CR )
Figure 4.6
Problem 20. The temperature θ 2 of a winding
which is being heated electrically at time t is given
by: θ 2 = θ 1 (1 − e
−t
τ ) where θ 1 is the temperature (in
degrees Celsius) at time t = 0 and τ is a constant.
Calculate,
(a) θ 1 , correct to the nearest degree, when θ 2 is
50 ◦ C, t is 30 s and τ is 60 s
(b) the time t , correct to 1 decimal place, for θ 2 to
be half the value of θ 1 .
(a) Transposing the formula to make θ 1 the subject
gives:
θ 1 =
θ 2
(1 − e
−t
T )
=
50
1 − e
−30
60
=
50
1 − e −0.5 =
50
0.393469 ...
i.e. θ 1 = 127
◦ C, correct to the nearest degree.
(b) Transposing to make t the subject of the formula
gives:
θ 2
θ 1
= 1 − e
−t
τ
from which, e
−t
τ = 1 −
θ 2
θ 1
Hence
−
t
τ
= ln
1 −
θ 2
θ 1
i.e.
t = −τ ln
1 −
θ 2
θ 1
Since
θ 2 =
1
2
θ 1
t = −60 ln
1 −
1
2
= −60 ln 0.5 = 41.59 s
Hence the time for the temperature θ 2 to be
one half of the value of θ 1 is 41.6 s, correct to 1
decimal place.
Now try the following exercise
Exercise 18 Further problems on the laws
of growth and decay
1. The temperature, T ◦ C, of a cooling object
varies with time, t minutes, according to the
equation: T = 150e −0.04t . Determine the temperature when (a) t = 0, (b) t = 10 minutes.
[(a) 150 ◦ C (b) 100.5 ◦ C ]
2. The pressure p pascals at height h metres
above ground level is given by p = p 0 e
−h
C ,
where p 0 is the pressure at ground level
and C is a constant. Find pressure p when
p 0 = 1.012 × 10 5 Pa, height h = 1420 m, and
C = 71500.
[99210]
3. The voltage drop, v volts, across an inductor L henrys at time t seconds is given
by v = 200 e
− Rt
L , where R = 150 and
L =12.5 × 10 −3 H. Determine (a) the voltage
when t = 160 ×10 −6 s, and (b) the time for the
voltage to reach 85 V.
[(a) 29.32 volts (b) 71.31 × 10 −6 s]
4. The length l metres of a metal bar at temperature t ◦ C is given by l = l 0 e αt , where
l 0 and α are constants. Determine (a) the
value of α when l = 1.993 m, l 0 = 1.894 m
and t = 250 ◦ C, and (b) the value of l 0 when
l = 2.416, t = 310 ◦ C and α = 1.682 ×10 −4 .
[(a) 2.038 × 10 −4 (b) 2.293 m]
5. The temperature θ
◦
2 C of an electrical conductor at time t seconds is given by:
θ 2 = θ 1 (1 − e −t / T ), where θ 1 is the initial
temperature and T seconds is a constant.
Determine:
(a) θ 2 when θ 1 = 159.9 ◦ C, t = 30 s and
T = 80 s, and
(b) the time t for θ 2 to fall to half the value
of θ 1 if T remains at 80 s.
[(a) 50 ◦ C (b) 55.45 s ]
6. A belt is in contact with a pulley for a
sector of θ = 1.12 radians and the coefficient